Question

A random sample of $100$likely voters in a small city produced $59$voters in favor of Candidate A. The observed value of the test statistic for testing the null hypothesis $H_0:p=0.5$versus the alternative hypothesis $H_a:p=0.5$is

$\text{(a)}z=\frac{0.59-0.5}{\sqrt{\frac{0.59(0.41)}{100}}}$

(b). $z=\frac{0.59-0.5}{\sqrt{\frac{0.5(0.5)}{100}}}$

(c). $z=\frac{0.5-0.59}{\sqrt{\frac{0.59(0.41)}{100}}}$

(d). $z=\frac{0.5-0.59}{\sqrt{\frac{0.5(0.5)}{100}}}$

(e).$t=\frac{0.59-0.5}{\sqrt{\frac{0.5(0.5)}{100}}}$

Short answer

The correct answer is (b)$z=\frac{0.59-0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}$.

Step-by-step solution

Given Information

$x=59$

$n=100$$H_0:p=0.5$

$H_a:p>0.5$

Explanation

The sample proportion is the number of successes divided by the sample size:

$\hat{p}=\frac{x}{n}=\frac{59}{100}=0.59$

Determine the value of the test-statistic:

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.59-0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}$